用户: Yao/泛函分析续论

Banach algebra	Banach space with $\Vert ab\Vert \le \Vert a\Vert \Vert b\Vert$
involution	$a^{\ast \ast }=a,(ab{)}^{\ast }=b^{\ast }{a}^{\ast },(\alpha a+b{)}^{\ast }=\bar{\alpha }{a}^{\ast }+b^{\ast }$
$C^{\ast }$ algebra	Banach algebra with $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$
hermitian elements $\mathrm{Re}\mathcal{A}$	$a^{\ast }=a$
maximal ideal space of abelian $\mathcal{A}$	$\Delta =\{$all nonzero homomorphisms $h:\mathcal{A}\to \mathbb{C}\}$
Gelfand transform of $a$	$\hat{a}:\Delta \to \mathbb{C},h\mapsto h(a)$
$C^{\ast }(a)$	the $C^{\ast }$ subalgebra generated by $a$ and $1$
positive elements in $\mathcal{A}$	$a\in \mathrm{Re}\mathcal{A}$ st $\sigma (a)\subseteq {\mathbb{R}}_{+}$
state space $\Sigma (\mathcal{A})$	all positive functionals of norm $1$

命题 0.1. The correspondence $h\mapsto \ker h$ between homopmorphisms and maximal ideals is bijective.

证明. For a maximal ideal $M$, let $\pi :\mathcal{A}\to \mathcal{A}/M$, and suppose that $\pi (a)$ is not invertible in $\mathcal{A}/M$. Note that $M\subseteq$ the ideal $I={\pi }^{-1}(\pi (⟨a⟩))={\pi }^{-1}(⟨\pi (a)⟩)$, then $M=I$, hence $\pi (⟨a⟩)=\pi (I)=0$, so $\pi (a)=0$. Since $\sigma (\pi (a))$ is nonempty, $\pi (a)$ is a number. Therefore $h:\mathcal{A}\to \mathcal{A}/M\simeq \mathbb{C}$ is such a homo that $\ker h=M$.

For a homo $h$, $A/\ker h\simeq \mathbb{C}$, so $\ker h$ is maximal.

$\square$

命题 0.2. $\sigma (a)=\{h(a):h\in \Sigma \}$.

证明. $a-h(a)\in \ker h$, so $h(a)\in \sigma (a)$; for $\lambda \in \sigma (a)$, a maximal ideal containing $a-\lambda$ is of the form $\ker h$, so $h(a)=\lambda$.

$\square$

命题 0.3. The kernel of $\Lambda :\mathcal{A}\to C(\Delta ),a\mapsto \hat{a}$ is the intersection of all maximal ideals of $\mathcal{A}$, and $\Vert \hat{a}{\Vert }_{\infty }=r(a)$.

证明. The former obvious, the latter by the former proposition.

$\square$

命题 0.4. Let $\mathcal{A}$ be a $C^{\ast }$ algebra.

(i)	$\Vert a\Vert =\Vert a^{\ast }\Vert$.
(ii)	For $a\in \mathrm{Re}\mathcal{A}$, $\Vert a\Vert =r(a)$.
(iii)	$\rho :\mathcal{A}\to \mathcal{B}$ is a $\ast$-homo, then $\Vert \rho (a)\Vert \le \Vert a\Vert$. (Coro: $\rho$ is $\ast$-iso implies isometry.)
(iv)	For unital abelian $\mathcal{A}$ and $h:\mathcal{A}\to \mathbb{C}$, $\Vert h\Vert =1$.
(v)	For $a\in \mathrm{Re}\mathcal{A}$, $h(a)\in \mathbb{R}$. (Coro: $h(a^{\ast })=\overline{h(a)}$.)
(vi)	If $\mathcal{A}$ is unital abelian, $\Lambda :\mathcal{A}\to C(\Delta )$ is a $\ast$-isomorphism. $\square$

证明.

(i)	$\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert \le \Vert a^{\ast }\Vert \Vert a\Vert$, and $a^{\ast \ast }=a$.
(ii)	$\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert =\Vert a^2\Vert$, so $r(a)=\underset{n\to \infty }{\lim }\Vert a^{2^n}{\Vert }^{1/2^n}=\Vert a\Vert$.
(iii)	By extending $\rho$ when necessary, we can assume $\mathcal{A}$ is unital and $\rho (1)=1$. Note that $\sigma (\rho (a))\subseteq \sigma (a)$, then $\Vert \rho (a){\Vert }^2=\Vert \rho (a^{\ast }a)\Vert =r(\rho (a^{\ast }a))\le r(a^{\ast }a)=\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.
(iv)	If there is $\Vert a\Vert /h(a)<1$, then exists $b=(1-a/h(a){)}^{-1}$, and it follows that $1=h(b(1-a/h(a))=0$.
(v)	Consider the abelian $C^{\ast }$ algebra generated by $a+it$ and $1$, where $t\in \mathbb{R}$. Then $$|h(a+it){|}^2\le \Vert a+it{\Vert }^2=\Vert a^2+t^2\Vert \le \Vert a^2\Vert +t^2.$$Write $h(a)=\alpha +i\beta$, then $|h(a+it){|}^2=|\alpha {|}^2+(\beta +t{)}^2$, forcing $\beta =0$.
(vi)	When $a\in \mathrm{Re}\mathcal{A}$, $\Vert \hat{a}{\Vert }_{\infty }=r(a)=\Vert a\Vert$. Hence for any $a$, $\Vert \hat{a}{\Vert }_{\infty }^2=\Vert {\hat{a}}^{\ast }\hat{a}{\Vert }_{\infty }=\Vert \widehat{a^{\ast }a}{\Vert }_{\infty }=\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2.$ Moreover, as $\Lambda (\mathcal{A})$ is a closed algebra separating points of $\Delta$, $\Lambda$ is surjective by Stone–Weierstrass. $\square$

定理 0.5.

(i)	For $a\in \mathrm{Re}\mathcal{A}$, $\sigma (a)\subseteq \mathbb{R}$.
(ii)	For two $C^{\ast }$ algebras $\mathcal{B}\subseteq \mathcal{A}$ with common identity, ${\sigma }_{\mathcal{B}}(a)={\sigma }_{\mathcal{A}}(a)$.

证明.

(i)

If $\mathcal{A}$ is abelian, it follows from (v) of the preceding proposition. If $\mathcal{A}$ is not, nonetheless $C^{\ast }(a)$ is abelian, then ${\sigma }_{\mathcal{A}}(a)\subseteq {\sigma }_{C^{\ast }(a)}(a)\subseteq \mathbb{R}$.

(ii)

When $a\in \mathrm{Re}\mathcal{B}$, ${\sigma }_{C^{\ast }(a)}\subseteq \mathbb{R}$ and $\partial {\sigma }_{C^{\ast }(a)}(a)\subseteq \partial {\sigma }_{\mathcal{A}}(a)$ implies that ${\sigma }_{\mathcal{B}}(a)={\sigma }_{\mathcal{A}}(a)$. Now for $a\in \mathcal{B}$, suppose that $a$ is invertible in $\mathcal{A}$, i.e. $ax=xa=1$, then $(a^{\ast }a)(x{x}^{\ast })=(x{x}^{\ast })(a^{\ast }a)=1$, where $a^{\ast }a\in \mathrm{Re}\mathcal{B}$ and is thus invertible in $\mathcal{B}$ (as we have shown in the previous step that ${\sigma }_{\mathcal{B}}(a^{\ast }a)={\sigma }_{\mathcal{A}}(a^{\ast }a)$), so $x{x}^{\ast }\in \mathcal{B}$ by the uniqueness of inverse, then $x=(x{x}^{\ast })a^{\ast }\in \mathcal{B}$. $\square$

定理 0.6. If $\mathcal{A}$ is a $C^{\ast }$ algebra, $a$ normal and $f\in C(\sigma (a))$, then $\sigma (f(a))=f(\sigma (a))$.

证明. The map $f\mapsto f(a)$ is a $\ast$-isomorphism from $C(\sigma (a))$ to $C^{\ast }(a)$, so $\sigma (f(a))={\sigma }_{C(\sigma (a))}(f)=f(\sigma (a))$.

$\square$

定理 0.7. Let $N$ be a normal operator, then there is a spectral measure $E$ defined on the Borel subsets of $\sigma (N)$ st for $f\in C(\sigma (N))$, $$f(N)={\int }_{\sigma (N)}zdE(z)$$with: (i) $\lambda$ is an eigenvalue of $N$ iff $E(\{\lambda \})\ne 0$; (ii) $TN=NT$ iff $TE(\omega )=E(\omega )T$ for all Borel $\omega$.

证明. (i) For eigenvalue $\lambda$, write $Nx=\lambda x$. Let $$f_n(z)=\{\begin{array}{ll}1/(\lambda -z),& z\notin B_{1/n}(\lambda ),\\ 0,& z\in B_{1/n}(\lambda ).\end{array}$$Then $$E(\mathbb{C}\setminus B_{1/n}(\lambda ))x=f_n(N)(\lambda -N)x=0,$$letting $n\to \infty$ gives $E(\mathbb{C}\setminus \{\lambda \})x=0$, i.e. $E(\{\lambda \})x=x$.

(ii) By Fuglede, $T{N}^{\ast }=N^{\ast }T$, so $Tp(N,N^{\ast })=p(N,N^{\ast })T$ for any polynomial, and subsequently $Tf(N)=f(N)T$ for any $f\in C(\sigma (N))$, and therefore $TE(\omega )=E(\omega )T$ for any open set $\omega$ by taking a sequence of nonnegative continuous $f_n\nearrow 1_{\omega }$, and ultimately for any Borel $\omega$. For the converse, approximate $z$ by simple functions.

$\square$

定理 0.8. Let $\mathcal{A}$ be a self–adjoint algebra of $B(H)$ with $\mathcal{A}v=0\Rightarrow v=0$. Let ${\mathcal{A}}_s$ and ${\mathcal{A}}_w$ be the strong and weak closures of $\mathcal{A}$. Then ${\mathcal{A}}_s={\mathcal{A}}_w={\mathcal{A}}^{\prime \prime }$.

证明. Clearly ${\mathcal{A}}_s\subseteq {\mathcal{A}}_w\subseteq {\mathcal{A}}^{\prime \prime }$, since if there is a sequence of $\{A_i\}$ in $\mathcal{A}$ st $⟨A_{i}h,g⟩\to ⟨Ah,g⟩$, then it follows from $⟨ASh,g⟩\leftarrow ⟨A_{i}Sh,g⟩=⟨S{A}_{i}h,g⟩\to ⟨SAh,g⟩$ that $AS=SA$ for all $S\in {\mathcal{A}}^{\prime }$. Then it suffices to show that every operator $B\in {\mathcal{A}}^{\prime \prime }$ can be strongly approximated by operatoes in $\mathcal{A}$; that is, to find $A\in \mathcal{A}$ st $\sum _{k=1}^n\Vert (B-A)h_k\Vert <\epsilon$.

Consider first the case $n=1$, and let $P$ be the projection onto $\overline{\mathcal{A}h}$. Since for all $A\in \mathcal{A}$ we have $A\overline{\mathcal{A}h}\subset \overline{\mathcal{A}h}$ and $A({\overline{\mathcal{A}h}}^{\perp })\subset {\overline{\mathcal{A}h}}^{\perp }$, $AP=PA$ so $P\in {\mathcal{A}}^{\prime }$.

$\square$